Indeed, let's introduce a second variable $x$ and consider the polynomial
$$
P = \sum_{i=0}^l y^i \cdot (1+x)^{l+i}
$$
By applying the binomial formula to $(1+x)^{l+i}$, we see that the coefficient of $x^l$ in $P$ is
$$
C_{l,l}(y) = \sum_{i=0}^l \binom{l+i}{l} y^i
$$

We now consider the twisted case
$$
C_{a,b}(y) \underset{def}{=} \sum_{i=0}^a \binom{b+i}{b} y^i.
$$
The method is exactly the same except that we compute the coefficient of $x^b$ in the polynomial
$$
\sum_{i=0}^a y^i (1+x)^{b+i}.
$$
We get
$$
C_{a,b}(y) = \frac{(1+\rho)^b}{1-y} - \frac{y^{a+1}}{1-y} \cdot \sum_{s=0}^b \binom{a+b+1}{b-s} \rho^s.
$$
Unfortunately, the last sum does not simplify as before because of the asymmetry $a \ne b$ (a priori).